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Question

# A hydrogen atom has an electrons in a particular excited state 'n', when it returns to the ground state, 6 different photons are emitted. Which of the following is/are incorrect:

A
Out of the 6 different photons only 2 photons are in the visible light region
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B
If highest energy proton emitted from the above sample is incident on the metal plate having work function 8 eV, KE of liberated photo-electron may be equal to or less than 4.75 eV
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C
Total number of radial nodes in all the orbitals of nth shell is 14
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D
Total number of angular nodes in all the orbitals of (n - 1) shell is 13
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Solution

## The correct option is D Total number of angular nodes in all the orbitals of (n - 1) shell is 13Number of photons emitted = 6 So, n × (n−1)2=6 n=4 So excited state is 3rd or n=4 Photon having highest energy will be 4→1 So its energy will be = 13.6× (112−142)=12.75 When it is incident on plate having work function 8 ev KE = 12.75 - 8 = 4.75 KE will be equal to this value or may be lower if the electron is an inner electron. So, option B is right, but we are looking for incorrect anwers. Option A is right too as two transitions happen in the Balmer series - 4→2, 3→2. C , D are incorrect because number of nodes in nth and (n - 1) shell are 6 (4s, 4p and 4d; 3+2+1+0 radial nodes) and 3 (3s, 3p, 3d; 2+1+0 Angular nodes) respectively.

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